Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-3x-4y &= 7 \\ -8x-8y &= 5\end{align*}$
Answer: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-8x = 8y+5$ Divide both sides by $-8$ to isolate $x$ $x = {-y - \dfrac{5}{8}}$ Substitute this expression for $x$ in the first equation. $-3({-y - \dfrac{5}{8}}) - 4y = 7$ $3y + \dfrac{15}{8} - 4y = 7$ Simplify by combining terms, then solve for $y$ $-1y + \dfrac{15}{8} = 7$ $-1y = \dfrac{41}{8}$ $y = -\dfrac{41}{8}$ Substitute $-\dfrac{41}{8}$ for $y$ in the top equation. $-3x-4( -\dfrac{41}{8}) = 7$ $-3x+\dfrac{41}{2} = 7$ $-3x = -\dfrac{27}{2}$ $x = \dfrac{9}{2}$ The solution is $\enspace x = \dfrac{9}{2}, \enspace y = -\dfrac{41}{8}$.